And I thought I understood stats ...

Hi all, new to the forums, new poster ...

I'm quite interested in the mathematics of gambling; and while surfing about I came across this entry in the gaming mathematics entry in wikipedia

Quote:

"Therefore, after 10 rounds betting $1 per round, your result will be somewhere between -$0.53 - 3 x $3.16 and -$0.53 + 3 x $3.16, i.e., between -$10.00 and $8.95. (There is still a 0.1% chance that your result will exceed a $8.95 profit, and a 0.1% chance that you will lose more than $10.00.)"

This was speaking about an even money bet on an American wheel. Now - I'm confused.

How can you possibly lose more than $10 betting $1 per round for 10 rounds? And isn't this interval a bit wide? Roulette, as I understand it is a fairly low volatility game. What am I missing?

Quote: alxhix

Hi all, new to the forums, new poster ...

I'm quite interested in the mathematics of gambling; and while surfing about I came across this entry in the gaming mathematics entry in wikipedia (http://en.wikipedia.org/wiki/Gaming_mathematics#Standard_deviation):

This was speaking about an even money bet on an American wheel. Now - I'm confused.

How can you possibly lose more than $10 betting $1 per round for 10 rounds? And isn't this interval a bit wide? Roulette, as I understand it is a fairly low volatility game. What am I missing?

I haven't read the article, but it appears to be a 99.8% confidence interval for a sample size of 10. I suspect it is a normal approximation to the binomial. The normal distribution doesn't know that its left hand tail has been truncated at -$10.00.

Quote: alxhix

How can you possibly lose more than $10 betting $1 per round for 10 rounds? What am I missing?

There is a 1/10 of 1% chance that you'll tip the drink girl during those 10 spins?

10 bets is really too small of a sample size to be applying the normal distribution. 25 is the minimum that I'd suggest. After 25 even money bets, you'll usually be up or down by 5 bets or less, and almost always be up or down by 15 or less.

Quote: PapaChubby

10 bets is really too small of a sample size to be applying the normal distribution.

Yes, very small sample size. This is the problem with math applied to events in the short run.
Remember, Binomial Standard Deviation deals with exact numbers, not exact possible outcomes as does the Binomial Distribution.

So you end up with a SD of 1.57.
Even the 5.26% HA can not be achieved in 10 trials.
The smaller chart inside the photo is a Binomial standard Deviation formula from the Wizard of Odds site. easier to use in Excel. I have learned some good math from this site. Cool stuff for me being 72 years old!

Quote: alxhix

Hi all, new to the forums, new poster ...

I'm quite interested in the mathematics of gambling; and while surfing about I came across this entry in the gaming mathematics entry in wikipedia (http://en.wikipedia.org/wiki/Gaming_mathematics#Standard_deviation):

This was speaking about an even money bet on an American wheel. Now - I'm confused.

How can you possibly lose more than $10 betting $1 per round for 10 rounds? And isn't this interval a bit wide? Roulette, as I understand it is a fairly low volatility game. What am I missing?

The math is correct, the problem is small sample size working with Binomial Std deviation. see my above post

Quote: PapaChubby

10 bets is really too small of a sample size to be applying the normal distribution. 25 is the minimum that I'd suggest. After 25 even money bets, you'll usually be up or down by 5 bets or less, and almost always be up or down by 15 or less.

The rule of thumb is usually np>5--so The use of the normal here is borderline. The continuity correction needs to be used also.

Cool - so it's not really the math that's an issue - it's simply a small sample, with an approximation that has no knowledge of its outcome set being truncated.

Makes sense I s'pose.

Thanks for the help.